Solve
$(3 x^{2} + 4 y) \times (2 x + 3 y)$

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Solution

We have,
$(3 x^{2} + 4 y) \times (2 x + 3 y)$
$(a + b) (c + d) = a c + a d + b c + b d$
$= 3 x^{2} .2 x + 3 x^{2} .3 y + 4 y .2 x + 4 y .3 y$
$= 3.2 x^{2} x + 3.3 x^{2} y .2 x y + 4.3 y y$
$6 x^{3} + 9 x^{2} y + 8 x y + 12 y^{2} = 0$

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Similar questions

2 x + 3 y = 11

2 x - 4 y = - 24

x + 3 y = 24 2 x + 4 y = 36

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\frac{2 x}{3} + 4 y + \frac{1}{2} = 0

2 x + 3 y = 11

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